A air-foce plane is ascending vertically at the rate of 100 km/h. If the radius of the earth is r km, how fast is the area of the earth, visible from the plane, increasing at 3 min after it started ascending?
Note Visible area `A=(2pir^(2)h)/(r+h)`, where h is the height of the plane above the earth.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Formula for Visible Area
The visible area \( A \) of the Earth from the plane is given by the formula:
\[ A = \frac{2 \pi r^2 h}{r + h} \]
where \( r \) is the radius of the Earth and \( h \) is the height of the plane above the Earth's surface.

### Step 2: Differentiate the Area with Respect to Time
We need to find how fast the area \( A \) is increasing, which means we need to differentiate \( A \) with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt} \left( \frac{2 \pi r^2 h}{r + h} \right) \]
Using the quotient rule, we differentiate:
\[ \frac{dA}{dt} = 2 \pi r^2 \cdot \frac{(r + h) \frac{dh}{dt} - h \frac{d}{dt}(r + h)}{(r + h)^2} \]
Since \( r \) is constant (the radius of the Earth), \( \frac{dr}{dt} = 0 \).

### Step 3: Simplify the Derivative
Substituting \( \frac{dr}{dt} = 0 \) into the equation gives:
\[ \frac{dA}{dt} = \frac{2 \pi r^2 \left( (r + h) \frac{dh}{dt} - h \cdot 0 \right)}{(r + h)^2} \]
This simplifies to:
\[ \frac{dA}{dt} = \frac{2 \pi r^2 (r + h) \frac{dh}{dt}}{(r + h)^2} \]

### Step 4: Calculate Height \( h \) After 3 Minutes
The plane ascends at a rate of \( 100 \) km/h. After \( 3 \) minutes, we convert minutes to hours:
\[ 3 \text{ minutes} = \frac{3}{60} \text{ hours} = \frac{1}{20} \text{ hours} \]
Now, calculate \( h \):
\[ h = \text{speed} \times \text{time} = 100 \text{ km/h} \times \frac{1}{20} \text{ h} = 5 \text{ km} \]

### Step 5: Substitute Values into the Derivative
Now we substitute \( h = 5 \) km and \( \frac{dh}{dt} = 100 \) km/h into the derivative:
\[ \frac{dA}{dt} = \frac{2 \pi r^2 (r + 5) \cdot 100}{(r + 5)^2} \]
This simplifies to:
\[ \frac{dA}{dt} = \frac{200 \pi r^2}{r + 5} \]

### Final Step: Conclusion
Thus, the rate at which the area of the Earth visible from the plane is increasing at 3 minutes after it started ascending is:
\[ \frac{dA}{dt} = \frac{200 \pi r^2}{r + 5} \text{ km}^2/\text{h} \]